(2014•虹口区一模)设函数fn(x)=−2n+2x+22x2+…+2nxn.
(2010•虹口区一模)已知函数f(x)=a2x+13x−1(a∈N),方程f(x)=-2x+7有两个根x1,x2,且x
(2014•嘉定区二模)设fn(x)=sin(nπ2
(2014•虹口区一模)函数f(x)=2sinπx与函数g(x)=3x−1
(2010•虹口区一模)已知二次函数y=x2+2x-3,解答下列问题:
设f(x)=–2x+2,记f1(x)=f(x),fn(x)=f[fn-1(x)],n≥2,n∈N,则函数y=fn(x)的
求和Sn=x+2x2+3x3+…+nxn(x≠0).
求和 l+2x+3x2+…+nxn-1.
(2010•虹口区一模)已知函数f(x)=sin2x−23cos2x+3,x∈[π4,π2].
(2013•虹口区一模)设点P在曲线y=x2+2上,点Q在曲线y=x−2
已知函数f1(x)=(2x-1)/(x+1) 对于n∈N* 定义fn+1(x)=f1( fn(x)) 求fn(x)解析式
(2014•虹口区一模)已知x:y=3:2,那么(x+y):x=______.
求和:Sn=1+2x+3x2+…+nxn-1.